2024 AMC 10A Problem 16

Let every rectangle have sides in the ratio $r : 1$ (long to short), the same $r$ for all since they are similar. A rectangle of area $S$ then has short side $\sqrt{S/r}$ and long side $\sqrt{Sr}$.

The figure is to scale, so orientations can be read off. Rectangle $8$ is wide (its width is its long side), while rectangles $9$ and $1$ are tall (their widths are short sides), and $8$ sits exactly on top of $9$ and $1$. Matching widths:
$$\sqrt{8r} = \sqrt{9/r} + \sqrt{1/r}.$$
Multiply through by $\sqrt r$: $r\sqrt8 = 3 + 1 = 4$, so $r = \dfrac{4}{2\sqrt2} = \sqrt2$.

(Check with another spot: tall $16$ sits above tall $4$ and wide $2$, so $\sqrt{16/r} = \sqrt{4/r} + \sqrt{2r}$, i.e. $4 = 2 + r\sqrt2$, again $r = \sqrt2$.)

The total area is $36+16+8+18+4+2+9+1+32+25+49 = 200$. The enclosing rectangle is wide, so its width $AB$ is the long side: $AB = \sqrt{200\,r} = \sqrt{200\sqrt2}$. Since $200\sqrt2 = 100\cdot 2^{3/2}$, we get $AB = 10\cdot 2^{3/4} = 10\sqrt[4]{8}$.

The answer is $\boxed{\textbf{(D)}\ 10\sqrt[4]{8}}$.

Similar figures have all lengths proportional to $\sqrt{\text{area}}$, so a single unknown ratio $r$ describes every rectangle; one place where a side is split by two neighbors gives an equation for $r$. Reading orientation from a to-scale figure is legitimate and necessary. Once $r$ is known, the total area finishes the job.

Eliminate: the enclosing rectangle has area $200$ and is visibly about $1.4$ times as wide as tall, so $AB \approx \sqrt{200\cdot 1.4} \approx 16.7$. Choices evaluate to about $12.9, 14.1, 16.2, 16.8, 20$; this rules out (A), (B), (E) and favors (D), but the exact computation is needed to separate (C) from (D).

• Assuming all rectangles share the same orientation; the figure is to scale, so a wide rectangle of area S has width √(Sr) and a tall one has width √(S/r).
• Treating the enclosing rectangle as a square ($AB = \sqrt{200} = 10\sqrt2$, choice (B)).